Can you some help me with this word problem with quadratic equation. I am stumped ???

Question

Explain the conditions under which a quadratic equation can be solved using simple rerrangement, versus the quadratic formula. Is there a way to decide, from the structure of the equation being solved, when each approach is appropriate?

Answer 1

If a quadratic equation is re-arranged so that it is in descending powers of x, with the right -hand side of the equation being 0, then inspection( the math word for eye-balling it) will often reveal what the factors are.
EXAMPLE: x^2 = x+6.
Rearrangement gives x^2 -x-6=0,and the factors are, by inspection, (x-3)(x+2)=0. Since either (x-3) or (x+2)
must equal 0 to satisfy the multiplication, we can conclude x=3 or x= -2
If we elect to solve using the quadratic formula, we would of course use x= {-b + or - rt(b^2-4ac)}/2a, where a= coefficient of the x^2 term, b of the x term, and c is the constant.
To decide which approach is appropriate, it is useful to simply calculate the value of the DISCRIMINANT, which is the value under the square root sign. That would be b^2 -4ac.
If it is negative, you are scuppered: you cannot find the square root of a negative number in the Real Number system.(you can using Complex Numbers, but that requires a whole new set of rules, an is beyond the scope of this discussion)
If it is 0, you have two roots, both the same. Basically one solution for x.
If the discriminant is positive, you will have two real , unequal roots. If in addition the discriminant is a perfect square, then you can be assured that factoring by inspection is possible. By then however, you have done a calculation that is close to the answer. Why go back to inspection?
The answer is that calculating the value of the discriminate often produces a value that is NOT a perfect square. In that case, carry on using the quadratic formula.

I hope this is useful to you.

Answer 2

1) The area of a square is (side)^2. x^2 = 121 sqrt (x^2) = sqrt 121 x = 11 2) The perimeter of a figure is just the sum of all its sides. (x+5) + (3x-8) + (x+3) + (x^2-3x) = 120 x^2 + 2x = 120 x^2 + 2x - 120 = 0 (x - 10)(x + 12) = 0 x = 10 or -12 -12 would give us some negative values for the length of the sides, so we can eliminate that one. x = 10 Now plug that into the equation for each side. x+5 = 10 + 5 = 15 3x - 8 = 3(10) - 8 = 22 x+3 = 10 + 3 = 13 x^2 - 3x = (10)^2 - 3(10) = 70 3) The area of a parallelogram is (base)(height). From your description, it sounds like the base is (x+5) and the height is (x-5). (x+5)(x-5) = 96 x^2 - 25 = 96 x^2 = 121 x = 11 so x+5 = 11+5 = 16 x-5 = 11-5 = 6

Answer 3

Let the equation be ax^2+bx+c = 0
If b^2-4ac = 0 there will be just one root = -b/2a
If b^2-4ac <0 then there are no real roots and so no real factors are possible.
If b^2-4ac>0 there will be two real roots. If b^2-4ac is a perfect square, there will be two rational roots.

With practice, you will be able to spot which approach to take.
x^2-5x+6 = (x-3)(x-2) because (-2)(-3)= 6 and -2x-3x = -5x.
Look at the factors of 6. They are 1,2,3,6. Look at the factors of the x^2 term. They are 1. So possiblle roots are 6/1,3/1,2/1,and 1/1. In this case it turned out to be 2and 3.

Answer 4

Yes. Not all. Some can be done only with a formula. Others like this:
Splitting the middle term method. The middle term should be devided into two pieces in such a way that their sum will be equal to the middle term and their product will be equal to the product of the extreme terms.
For eg.:
4x²+16x+16
Here, middle term is 16x. This can be devided like these:
1x+15x, 2x+14x, 3x+13x, 4x+12x, 5x+11x, 6x+10x, 7x+9x, 8x+8x
The extreme terms are 4x² and 16. and their product is 64x².
From the combinations of the middle term ,only 8x & 8x will give us a product 64x² .So the quadratic expression can be rewritten as:
4x²+8x+8x+16
Let us take the common factor of the first two terms and of the last two terms.
ie. 4x²+8x=4x(x+2); 8x+16= 8(x+2)
Let us combine both:
4x(x+2) + 8(x+2)
In the above, (x+2) is common
Hence, (x+2)[4x+8]
There fore the factors are: (x+2) and (4x+8)....Ans.
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Answer 5

form of a quadratic equation is:

ax^2 +b*x + c = 0 where a,b,c are constants, a isn't = 0

For it to be solved using simple rearrangement (ie presume not factorisation), b has to be 0. Thus equation becomes:

ax^2 + c =0, and x = +- (-c/a)^1/2.

(Another way of looking at is is that minimum point on quadratic always lies on y-axis, because b = 0.)